kmu 496 materials science and technology...

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KMU 496 MATERIALS SCIENCE AND

TECHNOLOGY III

April, 2019

CHAPTER 10SOLID SOLUTIONS AND PHASE DIAGRAMS

Dr. Damla ÇETİN ALTINDAL

Main objectives of this chapter are to explore

❑The formation of solid solutions;

❑The effects of solid-solution formation on themechanical properties of metallic materials;

❑The conditions under which solid solutions canform;

❑The development of some basic ideasconcerning phase diagrams and

❑The solidification process in simple alloys.

Outline

❑Phases and the Phase Diagram

❑Solubility and Solid Solutions

❑Conditions for Unlimited Solid Solubility

❑Solid-Solution Strengthening

❑ Isomorphous Phase Diagrams

❑Relationship Between Properties and thePhase Diagram

❑Solidification of a Solid-Solution Alloy

❑Nonequilibrium Solidification and Segregation

Section 10.1- Phases and ThePhase Diagram

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Phases and The Phase Diagram➢Pure metallic elements have engineering

applications; for example, ultra-high puritycopper (Cu) or aluminum (Al) is used to makemicroelectronic circuits.

➢In most applications, however, we use alloys.

➢We define an “alloy” as a material that exhibitsproperties of a metallic material and is madefrom multiple elements.

➢A plain carbon steel is an alloy of iron and carbon.

➢Corrosion-resistant stainless steels are alloys thatusually contain iron, carbon, chromium, nickeland some other elements.

➢There are alloys based on aluminum, copper,cobalt, nickel, titanium, zinc and zirconium.

There are two types of alloys:

➢single-phase alloys and multiple-phasealloys.

➢In this chapter, we will examine the behaviorsof single-phase alloys.

Phases and The Phase Diagram

Phase - Any portion of a system which is physicallyhomogeneous within it and bounded by a surfaceso that it is mechanically separable from anyother portions.

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➢A phase has the following characteristics:▪ the same structure or atomic arrangement

▪ the same composition and properties

▪ a definite interface between the phase andsurrounding

➢Water has three phases:▪ Liquid water

▪ Solid ice

▪ Steam


Phases and The Phase Diagram➢Gibbs phase rule:

It describes the relationship between the number ofcomponents and the number of phases for a given systemand the conditions that may be allowed to change (e.g.,temperature, pressure, etc.).

• C: the number of chemically independent components, usually elements or compounds, in the system

• F: the number of degrees of freedom, or the number of variables

• P: the number of phases

2 + C = F + P

➢Phases do not always have to be solid, liquid, and gaseous forms of a material.

– Iron (Fe), can exist in FCC and BCC crystalstructures. These two solid forms of iron are twodifferent phases of iron that will be stable atdifferent temperatures and pressure conditions.

– Carbon can exist in many forms (e.g., graphite ordiamond).


▪ P-T diagram: A diagram describing thermodynamicstability of phases under different temperature andpressure conditions.


2 + C = F + P

Point A

2 + 1 = F + 1

F=2

2 degrees of freedom

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Example 10.1 SOLUTION

- In space the pressure is very low. Even at relatively low temperatures, solidmagnesium can begin to change to a vapor, causing metal loss that could damage aspace vehicle.

- A low-density material with a higher boiling point (and, therefore, lower vaporpressure at any given temperature) might be a better choice.

- Other factors to consider: In load-bearing applications, we should not only lookfor density but also for relative strength. Therefore, the ratio of Young’s modulus todensity or yield strength to density could be a better parameter to comparedifferent materials.

Example 10.1

Design of an Aerospace Component

Because magnesium (Mg) is a low-density material (ρMg = 1.738g/cm3), it has been suggested for use in an aerospace vehicleintended to enter the outer space environment. Is this a gooddesign?

Section 10.2- Solubility and Solid Solutions

Solubility and Solid Solutions

➢ Solubility - The amount of one material thatwill completely dissolve in a second materialwithout creating a second phase.

▪ Unlimited solubility - When the amount ofone material that will dissolve in a secondmaterial without creating a second phaseis unlimited.

▪ Limited solubility - When only a maximumamount of a solute material can bedissolved in a solvent material.

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Polymeric Systems

➢We can process polymeric materials toenhance their usefulness, which is similar tothe formation of solid solutions in metallic andceramic systems.

➢We can form materials that are known ascopolymers that consist of differentmonomers.

Polymeric Systems

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➢ For example, acrylonitrile (A), butadiene (B), and styrene(S) monomers can form a copolymer known as ABS.

➢ This copolymer is similar to a solid solution in that it hasthe functionalities of the three monomers from which it isderived, blending their properties.

Section 10.3- Conditions for UnlimitedSolid Solubility

Conditions for Unlimited Solid Solubility


➢In order for an alloy system, such as copper-nickel to have unlimited solid solubility, certainconditions must be satisfied.

➢These conditions, the Hume-Rothery rules, areas follows:

▪ Size factor

▪ Crystal structure

▪ Valence

▪ Electronegativity

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o Size Factor: The atoms or ions must be of similar size,with no more than a 15% difference in atomic radius.

o Crystal structure: The materials must have the samecrystal structure; otherwise, there is some point atwhich a transition occurs from one phase to a secondphase with a different structure.

o Valence: The ions must have the same valence;otherwise, the valence electron difference encouragesthe formation of compounds rather than solutions.

o Electronegativity: The atoms must have approximatelythe same electronegativity.



Section 10.4- Solid Solution Strengthening

Solid-Solution Strengthening

❖Solid-solution strengthening - Increasingthe strength of a metallic material via theformation of a solid solution.

❖Jewelry could be made out from pure goldor silver; however, pure gold and puresilver are extremely soft.

❖Jewelers add copper to gold and silver sothat the jewelry will retain its shape.

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• Degree of solid-solution strengthening

Depends on two factors:

– a large difference in atomic size between theoriginal (host or solvent) atom and the added (guestor solute) atom increases the strengthening effect.• A larger size difference produces a greater disruption of

the initial crystal structure.

– the greater the amount of alloying element added,the greater the strengthening effect.

Solid-Solution Strengthening Effect of Solid-Solution Strengthening on Properties

1. The yield strength, tensile strength, and hardness of thealloy are greater than those of the pure metals.

2. Almost always, the ductility of the alloy is less than that ofthe pure metal.

NOTE: Only rarely, as in copper-zinc alloys, doessolid-solution strengthening increase both strengthand ductility.

3. Electrical conductivity of the alloy is much lower than thatof the pure metal. Therefore, solid-solution strengthening ofcopper or aluminum wires for transmission of electricalpower is not recommended.

4. The resistance to creep and strength at elevatedtemperatures is improved by solid solution strengthening.

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Section 10.5 - Isomorphous PhaseDiagrams

Isomorphous Phase Diagrams

❑ Phase diagram: A phase diagram shows the phasesand their compositions at any combination oftemperature and alloy composition.

❑ Isomorphous phase diagram: A phase diagram inwhich components display unlimited solid solubility.

❑ Binary phase diagram: A phase diagram for asystem with two components.

❑ Ternary phase diagram: A phase diagram for asystem with three components.


Cu-Niphase diagram


Liquidus temperature: The temperature at which thefirst solid begins to form during solidification.

The liquidus temperature is defined as thetemperature above which a material iscompletely liquid.

Solidus temperature: The temperature below whichall liquid has completely solidified.

The solidus temperature is the temperaturebelow which the alloy is completely solid.

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Cu-40% Ni alloy

Liquidus temperature1280°C

Solidus temperature1240°C

➢ The temperature difference between the liquidus andthe solidus is the freezing range of the alloy.

➢Within the freezing range, two phases coexist: a liquidand a solid.

Cu-40% Ni alloy

Freezing range1280 - 1240 = 40°C

Pure metals solidify at a fixed temperature

(i.e., the freezing range is zero degrees).

✓Often we are interested in which phases arepresent in an alloy at a particular temperature.

✓If we plan to make a casting, we must be surethat the metal is initially all liquid.

✓If we plan to heat an alloy component, we mustbe sure that no liquid forms during the process.

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➢For each phase, we can specify a composition,expressed as the percentage of each element inthe phase.

➢Usually the composition is expressed in weightpercent (wt%).

➢When only one phase is present in the alloy, thecomposition of the phase equals the overallcomposition of the material.

Compositions of Each Phase

➢We keep the pressure fixed at one atmosphere,which is normal for binary phase diagrams. Thephase rule can be rewritten as:

1 + C = F + P (for constant pressure)

C is the number of independent chemical components, P is the number of phases (not pressure),F is the number of degrees of freedom.

We now use the number 1 instead of the number 2because we are holding the pressure constant. Thisreduces the number of degrees of freedom by one.

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➢Because there is only one degree of freedom in atwo-phase region of a binary phase diagram, thecompositions of the two phases are always fixedwhen we specify the temperature.

➢We can use a tie line to determine thecomposition of the two phases.

➢A tie line is a horizontal line within a two-phaseregion drawn at the temperature of interest.

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Amount of Each Phase (the Lever Rule)

➢We are interested in the relative amounts ofeach phase present in the alloy.

➢These amounts are generally expressed asweight percent (wt %).

➢We express absolute amounts of differentphases in units of mass or weight (grams,kilograms, pounds, etc.).

Example 10.8 SOLUTION

Let’s say that x = mass fraction of the alloy that is solid α. Since wehave only two phases, the balance of nickel must be in the liquidphase (L). Thus, the mass fraction of nickel in liquid will be 1 - x.

Total mass of nickel in 100 grams of the alloy = mass of nickel inliquid + mass of nickel in α.

So,

100 (% Ni in alloy) = [(100)(1 – x)](% Ni in L) + (100)[x](% Ni in α)

x = (40-32)/(45-32) = 8/13 = 0.62

If we convert from mass fraction to mass percent, the alloy at1250°C contains 62% α and 38% L.

Note that the concentration of Ni in α phase (at 1250°C) is 45% andconcentration of nickel in liquid phase (at 1250°C) is 32%.

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➢The lever rule in general can be written as;

➢ Sometimes we wish to express composition as atomicpercent (at %) rather than weight percent (wt %).

➢ For a Cu-Ni alloy, where MCUand MNi are the molecularweights, the following equations provide examples formaking these conversions:

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